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Of the 20 lightbulbs in a box, 2 are defective. An inspector

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Of the 20 lightbulbs in a box, 2 are defective. An inspector [#permalink] New post 20 Jan 2016, 08:44
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Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?

Give your answer as a fraction.

Practice Questions
Question: 24
Page: 344
Difficulty: medium


[Reveal] Spoiler: OA
\(\frac{153}{190}\)

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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector [#permalink] New post 20 Jan 2016, 08:49
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Solution

The question is a bit tricky.

If you have to choose the NON defective light bulbs, then you must choose among 18 good one.

So, you do have \(\frac{18}{20} * \frac{17}{19} =\frac{153}{190}\)

The correct answer is \(\frac{153}{190}\)
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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector [#permalink] New post 23 Mar 2018, 13:02
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Expert's post
Carcass wrote:
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability
that neither of the lightbulbs selected will be defective?

Give your answer as a fraction.

Practice Questions
Question: 24
Page: 344
Difficulty: medium


[Reveal] Spoiler: OA
\(\frac{153}{190}\)


P(neither lightbulb is defective) = P(BOTH lightbulbs work)
= P(1st lightbulb works AND 2nd lightbulb works)
= P(1st lightbulb works) x P(2nd lightbulb works)
= 18/20 x 17/19
= 9/10 x 17/19 [since 18/20 = 9/10]
= 153/190

Answer: 153/190

ASIDE: P(1st lightbulb works) = 18/20, because there are 20 lightbulbs and 18 of them work.
P(2nd lightbulb works) = 17/19 because, once we have selected a working bulb for the 1st selection, there are 19 lightbulbs remaining and 17 of them work.

Cheers,
Brent
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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector [#permalink] New post 08 Dec 2018, 19:51
I'm trying to solve this question using the complement method. P(defective) = 1 - P(NOT defective)

P(defective) = 2/20 * 1/19 = 1/190

Inserting in formula :

1/190 = 1 - P(NOT defective)
P(Not defective) = 189/190

What is wrong with this approach?
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Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector [#permalink] New post 09 Dec 2018, 00:48
Expert's post
oqureshi wrote:
I'm trying to solve this question using the complement method. P(defective) = 1 - P(NOT defective)

P(defective) = 2/20 * 1/19 = 1/190

Inserting in formula :

1/190 = 1 - P(NOT defective)
P(Not defective) = 189/190

What is wrong with this approach?



YOu are going wrong because the answer you are getting is for a different combination..
1-P(both defective) will give you Probability when both are not defective, so it will contain ways when one is defective and one is not AND this i snot the same as the probability
that neither of the lightbulbs selected will be defective?
..
so your complement method should be \(1-P(both -not- defective)-P(just- one -is -defective)=1-\frac{2*1}{20*19}-2*\frac{2*18}{20*19}=1-\frac{1}{190}-\frac{36}{190}=1-\frac{37}{190}=\frac{153}{190}\)..
You are multiplying just one defective by two as it can be defective first and then non-defective or non-defective first and then defective.
Of course here straight method is easier
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: Of the 20 lightbulbs in a box, 2 are defective. An inspector   [#permalink] 09 Dec 2018, 00:48
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